HAL Id: hal-00445256
https://hal.archives-ouvertes.fr/hal-00445256
Preprint submitted on 8 Jan 2010
HAL
is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire
HAL, estdestinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
Piecewise Certificates of Positivity for matrix polynomials
Ronan Quarez
To cite this version:
Ronan Quarez. Piecewise Certificates of Positivity for matrix polynomials. 2010. �hal-00445256�
POLYNOMIALS
RONAN QUAREZ
Abstract. We show that any symmetric positive definite homogeneous ma- trix polynomialM ∈ R[x1, . . . , xn]m×m admits a piecewise semi-certificate, i.e. a collection of identitesM(x) =P
jfi,j(x)Ui,j(x)TUi,j(x) whereUi,j(x) is a matrix polynomial andfi,j(x) is a non negative polynomial on a semi- algebraic subsetSi, whereRn=∪r
i=1Si. This result generalizes to the setting of biforms.
Some examples of certificates are given and among others, we study a vari- ation around the Choi counterexample of a positive semi-definite biquadratic form which is not a sum of squares. As a byproduct we give a representation of the famous non negative sum of squares polynomialx4z2+z4y2+y4x2− 3x2y2z2as the determinant of a positive semi-definite quadratic matrix poly- nomial.
1. Introduction
The Hilbert 17-th problem asks if a non negative polynomialf(x)∈R[x1, . . . , xn] is a sum of squares. The answer given by Artin provides an identity with denomi- nators (or rational functions), namely there are two sums of squares of polynomials q(x) andp(x) such thatq(x)f(x) =p(x).
Such an identity is a called Positivestellensatz and can be seen as a certificate which algebraically proves the positivity of the polynomial f(x). For a general polynomial, the denominatorq(x) is necessary.
One may also consider a ”relative” version, namely whenf(x) is non negative on a basic semi-algebraic subsetS={x∈Rn |f1(x)≥0, . . . , fr(x)≥0}. This search of certificates received a lot of contributions especially for those semi-algebraic subsetsSwhere the Positivestellensatz is denominator free. Essentially it happens to be possible when f(x) is positive and when the description ofS satisfies some archimedean property (which can be seen as a “strong” compactedness assumption).
Among all the available references, we mention the seminal works of Schm¨udgen [10] and Putinar [8].
Another natural and often difficult question is, when it is possible to obtain a certificate of positivity, to study its complexity, i.e. the number of squares that are needed.
In this paper, we are mainly interested in matrix polynomials, namely polyno- mials whose coefficients are matrices. Letx= (x1, . . . , xn) andR[x]m×mbe the set
Date: January 8, 2010.
2000Mathematics Subject Classification. 14P ; 15A.
Key words and phrases. Biforms ; Matrix polynomials ; Positive semi-definite ; Positivity certificate ; Sum of Squares.
1
of all square matrix polynomials of size m with entries in R[x]. In that context, by a square we mean a expression of the formU(x)TU(x) whereU(x)∈R[x]m×m, sometimes we will also use the terminology ofhermitiansquare.
Given a symmetric matrix polynomialM(x), the natural question is the following :
Assume thatM(x) is positive semi-definite (psdin short) for allx∈Rn. IsM(x) a sum of hermitian squares ? Or in otherwords, do we have a positivity certificate forM(x) ?
It is well known that, whenn= 1, then M(x) is a sum of two squares (see for instance [4] for a proof), i.e. M(x) =U(x)TU(x) +V(x)TV(x) whereU(x), V(x)∈ R[x]m×m.
There is also another case where we have a positive answer : the case of bi- quadratic forms in dimension 2. Namely, ifm= 2 and if the entries are homoge- neous polynomials of degree 2, thenM(x) is also a sum of squares.
In fact this last case can be seen as a particular case of the homogenized version of the previous one (see [2] for this fact and a proof).
Unfortunaltely these are essentially the only cases where such a certificate exists for a matrix polynomial. For instance, it becames false even for biquadratic forms in 3 variables and dimension 3 (confer the Choi counterexample in [1]).
We may also look at the relative situation : if M(x) is positive definite for all x in a semi-algebaric subset S ⊂ Rn whose description satisfies an archimedean property, then we get a similar certificate as in the polynomial case. This has been proved by Hol and Scherer in [6] ; see also [7] for a different and more algebraic proof.
In this paper, we are searching for certificates of positivitywithout denominators for matrix polynomials. We first study certificates which are sums of squares of matrix polynomial weighted by a psd (scalar) polynomial. More precisely, given a psd matrix polynomialM(x), we wonder ifM(x) can be written as a finite sum
(1) M(x) =X
i
fi(x)Ui(x)TUi(x)
whereUi(x)∈R[x]m×m andfi(x) is non negative polynomial onRn.
We call it asemi-certificatesince, roughly speaking, we have a certificate which is a hermitian sum of squares weighted by some coefficients which are non negative polynomials.
Of course such a certificate may gives some others which will be totally alge- braic by using usual certificates for the psd polynomials fi(x). Beware that using denominators at this point could break any interest since we may readily get one such certificate with denominators simply by using the Gauss algorithm (indeed, it suffices to inverte the m−1 first principal minors of the given matrix poly- nomial). Although, whene the matrix polynomial is positive definite, a relative version of semi-certificate with respect to a semi-algebraic subsetS whose descip- tion is archimedean would produce a usual denominator free algebraic certificate for the matrix polynomial on S, using for instance Schm¨udgen or Putinar certificates for all thefi,j’s.
Unfortunately such a semi-certificate as in (1) does not exist in general, even for a positive definite matrix polynomial, as it is illustrated in section 3.
Nevertheless, noticing that one may obtain such certificate locally, we introduce the notion ofpiecewisesemi-certificate, i.e. a collection of semi-certificates as in (1) with respect to a semi-algebraic covering ofRn.
We show in section 4, that such a certificate exists for any positive definite homogeneous matrix polynomial. We show also that we may translate the notion of semi-certificate and our related results to the case of biforms.
Some examples and counterexamples are given and for instance, in section 5, we study some variations around the Choi counterexample given in [1].
As a byproduct we give a representation of the famous psd non sum of squares polynomial x4z2+z4y2+y4x2−3x2y2z2 as the determinant of a psd quadratic matrix polynomial.
Of course, a lot a work have to be done to better understand the framework of semi-certificate for psd non definite matrix polynomials. Moreover, concerning complexity, one may be interested in estimating the lenght of the hermitian sum of squares representation, and also in the number of pieces of the semi-certificate.
Since we use a compactedness argument, no bound on the number of pieces of the semi-algebraic covering can be derived by our method.
2. Notations
2.1. Matrix polynomials and biforms. In all the paper, we consider a set of variables x= (x1, . . . , xn) and the polynomial ringR[x] whose associated field of rational functions isR(x).
A matrix polynomial A(x) is just a matrix whose entries are polynomials. For instance, for a square matrix polynomial of size m in the variables x, we write A(x)∈R[x]m×m.
A formof type (n, d) will denote an homogeneous polynomial of degree din n variables. Abiform of type (n1, d1;n2, d2) will denote a homogeneous polynomial in n1+n2 variables which is of degreed1 with respect to a set ofn1 variables and of degreed2 with respect to the other set ofd2 variables.
To a matrix polynomialA(x) = (ai,j(x))∈R[x]m×m whose entries are all forms of type (n, d), we may canonically associate a biform of type (n1, d1;m,2) :
fA(x, y) = X
1≤i,j≤m
ai,j(x)yiyj.
We say that a biform of type (n1, d1;n2, d2) ispositive definiteiff(x, y)>0 for all (x, y)∈Sn1−1×Sn2−1, whereSk−1 denotes the k1 dimentional unit sphere in Rk. Equivalentely, we may alternatively consider the product of projective spaces instead of unit spheres.
IfA(x) is a symmetric matrix polynomial inR[x]m×mwhose entries are all forms of type (n, d), we say thatA(x) is positive definite if the associated biformfA(x, y) is positive definite. We may also note thatA(x) is psd if and only iffA(x, y) is psd on the product of unit spheres.
2.2. Sum of squares and Hilbert 17th problem. Note that sums of hermitian squares when dealing with matrix polynomials corresponds to usual sum of squares in the terminology of biforms.
LetA(x) be a symmetric matrix polynomial inR[x]m×m which is positive semi- definite for all substitution ofx∈Rn. Then, the Gauss reduction algorithm gives a solution with denominators to the Hilbert 17-th Problem for a generic matrix poly- nomial : there existr∈N,U1(x), . . . , Ur(x)∈R[x]m×mandf1(x). . . , fr(x), f(x)∈ R[x] psd polynomials such that
(2) f(x)A(x) =
Xr i=1
fi(x)Ui(x)TUi(x)
wheref(x) can be choosen as the product of them−1 first minors ofA(x) when they do not identically vanish. Moreover, by multiplying the identity (2) by the denomiators appearing in a positivity certificate for thefi’s, we may assume that the polynomialsfi andf are sum of squares.
In our paper, we are mainly interested in positivity certificates without denom- inators, namely when we may choose f ≡ 1. Hence, we first look for matrix polynomials which can be written as in (1) :
A(x) = Xr i=1
fi(x)Ui(x)TUi(x)
where thefi’s are non-negative polynomials (we cannot ask for them to be sum of squares since denominators are already needed for polynomials).
One may want to measure the complexity of the certificate given in (1), simply by counting the number of squares : r. But, one may also prefer another way of counting.
Noticing thatUTU =UT(E1+. . .+Em)U = (E1U)T(E1U)+. . .+(EmU)T(EmU) where Ei is the diagonal matrix whose only non null entry is 1 at the i-th posi- tion onto the diagonial, we may prefer to count “rank-one” sums of squares, i.e.
whenU(x) has rank one inR(x)m×m. The advantage of that count is that it coin- cides with the number of needed squares when we view our matrix polynomial as a biform.
3. Semi-certificates
The following examples shows that it is not possible in general to hope for (1).
3.1. Some examples.
Example 3.1. Let :
M =
1 +x2 xy xy x2+y4
Of courseM is psd for all (x, y)∈R2.
To show that M cannot be written as in (1), we may proceed by applying to the biform fM which is canonically associated toM) the technology of cages and Gram matrices developped in [3]. But we prefer to produce here an elementary and self-contained argument.
Assume thatM =Pr
i=1fiUiTUi, whereUi∈(R[x, y])2×2 andfi≥0 on allR2.
By decomposing into rank one sum of squares, we may assume that UiTUi=
a2i aibi
aibi b2i
, whereai, bi∈R[x, y].
Identifying the (1,1)-entry, we getPr
i=1fia2i = 1 +x2. Thus,fi does not depend on y : fi ∈ R[x]. Since fi ≥0 for all x∈ R, we deduce thatfi is a sum of two squares in R[x]. Thus, if we increase the integer r and if we change the Ui’s, we may assume thatfi= 1. Identifying the entries, we get
Pr
i=1a2i = 1 +x2 Pr
i=1b2i = x2+y4 Pr
i=1aibi = xy
By a simple degree consideration, the polynomialbi has necessarily the form bi=αix+βiy2, withαi, βi∈R.
Likewise, the polynomialai has necessarily the form ai=γi+δixwithγi, δi∈R. It follows a contradiction with the identityPr
i=1aibi=xy.
The previous pathology is not due to the fact that the matrixM is only psd, as shown by considering the following variation :
Example 3.2. Let N =
1 +x2+ǫ(x4+y4) xy
xy ǫ(1 +x4) +x2+y4
where ǫ > 0. This is a positive definite matrix polynomial on all R2 (and even at infinity, i.e. the associated homogeneized matrix polynomial remains positive definite).
Assume thatN can be written as in (1) :
N =
Xr i=1
fi(x, y)
A2i AiBi
AiBi Bi2
+ Xr j=1
gj(x, y)
Uj2 UjVj UjVj Vj2
+ Pr k=1
R2k RkSk
RkSk Sk2
where the fi’s are polynomials of degree at most 4, the gj, Rk, Sk’s are polyno- mials of degree at most 2, the Uj, Vj’s are polynomials of degree at most 1 and theAi, Bi’s are constant. Moreover, the polynomialsfi, gj are assumed to be non negative onR2. Since their degrees are at most 4 and the number of variables is 2, we know that they are sum of squares of polynomials. Hence, we may assume that
(3) N =
Xr i=1
Ri2 RiSi RiSi Si2
where
Ri=ai+bix+cix2+diy+eiy2+fixy, Si=αi+βix+γix2+δiy+νiy2+µixy, (ai, bi, ci, di, ei, fi, αi, βi, γi, δi, νi, µi)∈R12.
Let us consider the following vectors in the usual euclidien spaceRr :
¯
a= (ai),¯b= (bi),c¯= (ci),d¯= (di),e¯= (ei),f¯= (fi),
¯
α= (αi),β¯= (βi),¯γ= (γi),δ¯= (δi),ν¯= (νi),µ¯= (µi).
Identifying the non diagonal entries in (3), we get (4) ¯b·¯δ+ ¯β·d¯+ ¯a·µ¯+ ¯f·α¯ = 1
where·denotes the usual inner product onRr.
By identifying the diagonal entries in (3) and by the Cauchy-Schwarz inequality, we get
¯b2 = 1−2 ¯a·c¯ ≤ 1 + 2√
¯ a2√
¯
c2 ≤ 1 + 2√ ǫ
¯δ2 = −2 ¯α·ν¯ ≤ 2√
¯ α2√
¯
ν2 ≤ 2√
¯ ǫ
β2 = 1−2 ¯α·γ¯ ≤ 1 + 2ǫ d¯2 = −2 ¯a·c¯ ≤ 2√
ǫ
¯
a2 = 1
¯
µ2 = −2 ¯γ·ν¯ ≤ 2√ ǫ f¯2 = −2 ¯c·¯e ≤ 2√
¯ c2√
¯
e2 ≤ 2ǫ
¯
α2 = ǫ
Again, by the Cauchy-Schwarz inequality, we have :
¯b·δ¯ ≤ p
2(1 + 2√ ǫ)√
ǫ β¯·d¯ ≤ p
2(1 + 2ǫ)√ ǫ
¯
a·µ¯ ≤ p 2√
ǫ f¯·α¯ ≤ √
2ǫ
Thus, if we takeǫsmall enough, then we get a contradiction to (4). Namely,N does not admit a certificate as in (1).
3.2. Semi-algebraic covering. Thus, even for positive definite matrix polynomi- als, a certificate (1) does not exist in general. Although, one may remark that such a certificate existslocallyin our examples.
Look at example 3.1. When|x| ≥ |y|, then consider the identity 1 +x2 xy
xy x2+y4
=
x2 xy xy y2
+
1 0 0 0
+ (x2−y2+y4)
0 0 0 1
whereas when|y| ≥ |x|, then consider
1 +x2 xy xy x2+y4
=
1 xy xy x2y2
+x2
1 0 0 1
+ (y4−x2y2) 0 0
0 1
Hence we are going to change our definition by considering a semi-algebraic covering of our space.
From now on, we will focus onforms, namelyhomogeneous polynomials.
Definition 3.3. Let f(x, y)be a biform of type(n, d1;m, d2). A semi-certificate of positivity is the data of a semi-algebraic covering Rn =S1∪. . .∪Sr such that
∀x∈Si, f(x, y) =
ri
X
j=1
fi,j(x)(gj,i(x, y))2
where eachfi,j(x)is a form of degreeei,j which is non negative onSi andgi,j2 (x, y) is a biform of type (n, d1−ei,j;m, d2).
Roughly speaking, the semi-certificate is a piecewise identity which is a sum of squares with respect to one set of variablesyand only psd with respect to the other set of variablesx.
For convenience, we may formulate what happens when we are dealing with matrix polynomials. The symmetric psd matrix polynomial M(x) ∈ R[x]m×m admits a semi-certificate if there exists a (finite) semi-algebraic covering (Si)1≤i≤r
ofRn and
(∀i)(∃ri∈N)(∀j∈ {1, . . . , si})(∃fi,j(x)∈R[x])(∃Ui,j(x)∈R[x]m×m) such that
(5)
M(x) =
ri
X
j=1
fi,j(x)Ui,jT (x)Ui,j(x)
and (∀x∈Si, fi,j(x)≥0)
In example 3.1, we have seen that it is possible to produce semi-certificate of positivity, where the considered semi-algebraic covering ofR2necessarily have more than one piece.
Remark 3.4. Note that if we restrict ourselves to certificates where theUi,j(x) are constant matrices, then we a strictly smaller class of certificates. For instance, consider the matrix polynomial
x2 xy xy y2
which is psd on allR2, which is obvi- ously psd on a neighbourhoodV of the point (1,1) ( we may proceed likewise the neighbourhood of any given point).
Let us deshomogenize by setting y= 1, set for simplicity V = [1,1 +ǫ[, ǫ >0, and assume that
x2 x x 1
=X
i
fi(x)
α2i αi
αi 1
+X
j
gj(x) 1 0
0 0
where thefi’s andgj’s are univariate polynomials psd onV andαi∈R.
Let
fi(x) =ai+bi(x−1) +ci(x−1)2 gj(x) =dj+ej(x−1) +fj(x−1)2
Then
x2 = P
iα2ifi(x) +P
jgj(x)
x = P
iαifi(x)
1 = P
ifi Combining these three identities we get
X
i
(αi−1)2fi(x) +X
j
gj(x) = (x−1)2
By assumption, we have ai ≥0 and dj ≥0. Then, for all j we havedj = 0 and ej = 0. Furthermore, ifαi6= 1, then ai= 0 and also bi = 0. But substracting the last two equalities of the previous system yields
x−1 =X
i
(αi−1)fi(x)
a contradiction.
4. Main results
There is one case when semi-certificates exist : when the matrix polynomial is positive definite. Remind that it means that it is positive definite on the unit sphere or the associated projective space.
Theorem 4.1. Let A(x) ∈ (R[x])m×m be a positive definite marix polynomial.
Then, there is a finite semi-algebraic covering ofRn=∪ri=1Si, some formsfi,j(x)∈ R[x], some matrix polynomialsAi,j(x)∈(R[x])m×m such that,
∀x∈Si, A(x) =
ri
X
j=1
fi,j(x)ATi,j(x)Ai,j(x) with the condition thatfi,j(x)≥0for all x∈Si.
Proof. Assume that the entries ofA(x) ared-forms (dis even).
We start with a lemma
Lemma 4.2. Let a(x) be a positive definite d-form and c(x) be a positive semi- definite d-form. Then, there existsǫ > 0 such that a(x)−ǫc(x) remains positive definite.
Proof. Thed-form ec(x) =c(x) +xd1+. . .+xdn is positive definite and the subset Sc˜ = {x | ˜c(x) = 1} is compact. Since a(x) > 0 on Sc˜ we have for all x ∈ S˜c, a(x)≥m >0. Hence, for allx,
a x
pd
˜
c(x), x pd
˜ c(x)
!
≥m >0.
Then a(x)− m2˜c(x) is positive definite, which concludes the proof since ˜c(x) ≥
c(x).
Let A(x) = (ai,j(x))1≤i,j≤m. For a given x0 ∈ Rn\ {0}, the matrix A(x) is positive definite in a semi-algebraic subset Ux0 which is open in the unit sphere Sn−1of Rn.
LetU be the square matrix of sizemwhose all entries are equal to 1. By 4.2 and up to resizingUx0, we may assume thatB(x) =A(x)−ǫ(xd1+. . .+xdn)U remains positive definite in Ux0 forǫsmall enough, and moreover that all entries ofB(x0) are non-zero.
LetB(x) = (bi,j(x))1≤i,j≤m. Assume thatb2,1(x0)>0. We may writeB(x0) = b2,1(x0)C where C ∈Rm×m is positive definite. Lethx0(x) be ad-form satisfying hx0(x0) = 1. Then, by 4.2,
b2,1(x)C−ǫ1hx0(x)Idm
remains positive definite for some smallǫ1>0.
Now write
B(x) = (b2,1(x)C−ǫ1hx0(x)Idm) +B(x)e
SinceB(xe 0) =ǫ1Idm, the matrixB(x) is positive definite in an open semi-algebraice neighbourhood ofx0which we still denote byUx0.
Then,B(x)) = (˜e bi,j(x))1≤i,j≤m is such that ˜b2,1(x) = 0.
Of course, we proceed likewise when b2,1(x0)<0, writingB(x0) =−b2,1(x0)C whereC∈Rm×mis positive definite. Thus,
B(x) = (−b2,1(x)C−ǫ1hx0(x)Idm) +B(x)e
Then, we repeat the same process to get rid off all thebi,j(x)’s such thati6=j and reduce, up to resizeUx0, to the case where B(x) is diagonal. Namely, there is an open semi-algebraic subsetUx0 ⊂Sn−1 such that for allx∈Ux0,
(6) A(x) =X
k
fk(x)UkTUk
whereUk ⊂Rm×mis a constant matrix andfk(x) is ad-form which is psd onUx0. To conclude the proof, we extract a finite semi-algebraic covering by compact- edness ofSn−1.
By homogeneity, we extend the identities (6) to allx∈Rn. Moreover, we may manage in order that the polynomials describing the Ux0’s are forms, and hence the identity (6) is true onSx0, the cone in Rn with origin 0 and basis Ux0. Thus,
we get a finite covering ofRn.
Note that the proof gives anopensemi-algebraic covering, and that theUk’s are constant matrices.
Likewise, we have an analogeous result for biforms :
Theorem 4.3. Letf(x, y)∈R[x, y]be a positive definite biform of type(n, d1;m, d2).
Then, there is a finite semi-algebraic covering ofRn=∪ri=1Si, some formsfi,j(x)∈ R[x], some biformsgi,j(x)∈R[x, y] such that,
∀x∈Si, f(x, y) =
ri
X
j=1
fi,j(x)(gi,j(x, y))2 with the condition thatfi,j(x)≥0for all x∈Si.
Proof. We proceed as in the proof of Theorem 4.1. We get rid of any monomial appearing inf with some odd power with respect to at least one indeterminate.
5. Semi-certificates on orthant-neighbourhoods
We say thatV is anorthant-neighbourhoodofx0if it contains the intersection of a neighbourhood ofx0and an orthantOx0 centered atx0(i.e. a subset defined by an open condition x=x0+ (X1, . . . , Xn)∈Ox0 if ǫ1X1>0, . . . , ǫnXn >0 where (ǫ1, . . . , ǫn)∈ {−1,+1}n). We will write thatOx0 is an orthant-neighbourhood of (x0, ǫ)
The orthant-neighbourhoods fit naturally with the use of Taylor expansion for- mula. Let us recall it relatively to an orthant-neigbourhood (x0, ǫ) for a matrix polynomialA(x) whose entries are not necessarily homogeneous polynomials :
A(x0+ǫX) =X
α
(ǫX)α
α! A(α)(x0)
where we use the standart multi-index symbol for products. Under a condition of domination by the constant term, we will see how to derive some certificate on orthant-neighbourhoods.
Definition 5.1. Let A(x) be a matrix polynomial. Denote by Γx0 the set of all multi-indexes β which are minimal (for the lexicographic ordering) such that A(β)(x0)6= 0.
We say thatA(x)satisfies the domination condition atx0 if for all multi-index αthere is some multi-indexβ ∈Γx0 and a non negative real numberrα,β such that β≤αandrα,βA(β)(x0)±A(α)(x0)is psd.
Note that Γx0={0}whenA(x0)6= 0 which simplifies the domination condition.
Note also that ifβ ∈Γx0 then necessarily all its coordinates are even integers.
Finally, mention that to check positivity via the Taylor formula, it would be enough to have the domination condition for any multi-indexαwhose at least one component is odd.
A typical use of this domination condition appears in the following situation : Proposition 5.2. Let A(x) be a psd matrix polynomial on a neighbourhood of x0. Then, A(x) satisfies the domination condition at x0 if and only if it admits a semi-certificate atx0 where theUi,j(x)’s appearing in (5) are constant matrices.
Proof. Assume that A(x) satisfies the domination condition at x0. Let V be an orthant-neighbourhood of (x0, ǫ). We have
A(x0+ǫX) = P
β∈Γx0
P
α∈∆β∪{β}
(ǫX)α
α! A(α)(x0)
where ∆β is a subset of all multi-index αsuch thatα > β (we have to be carefull that one multi-indexαmay be greater than several elements of Γx0).
A(x0+ǫX) = P
β∈Γx0
ǫβ Xβ!βA(β)(x0) 1−P
∆βǫβ β!α!Xα−βrα,β
+P
∆β
Xα
α! ǫαA(α)(x0) +rα,βA(β)(x0)
Sinceǫβ= 1, we obtain a semi-certificate on the orthant-neighbourhoodV. Conversely, assume that
A(x) =X
i
fi(x)Vi
where eachViis a constant psd matrix and eachfiis non negative on a neighbour- hood ofx0. We write the Taylor expansion offi atx0 :
fi(x0+X)Vi= X
α
Xα
α! fi(α)(x0)
!
×Vi
For the matrix polynomial fi(x)Vi, consider the set Γx0. If β ∈ Γx0, then fi(β)(x0) > 0 and for any α > β, we clearly have the existence of a positive real number rα,β such that rα,βfi(β)(x0)±fi(α)(x0)≥0. This is the domination condi-
tion.
Remark 5.3. Since positive definite matrices obviously satisfy the domination con- dition, we may recover a version of Theorem 4.1 with orthant-neighbourhoods. But, maybe (highly heuristic !) it will produce a lot more pieces for the covering, since (again roughly speaking) we need 2n−1 orthant-neighourhoods to recover a usual neighbourhood.
5.1. Semi-certificates relative to a semi-algebraic subset. One may natu- rally want to extend the framework of semi-certificates relatively to a basic closed semi-algebraic subset S. The problem is that the result given in this section does not take into account the equations describing S. We mainly use the underlying semi-algebraic set rather than the preordering or the quadratic module generated by the equations ofS as it is desired for a relative Positivestellensatz. In fact, this section concerns more the study oflocalsemi-certificates rather thanrelative’s ones.
Theorem 5.4. Let A(x) ∈ R[x]m×m be a homogeneous symmetric matrix poly- nomial. Assume that A(x) is positive definite for all x∈ S, where S is a closed semi-algebraically subset of Rn defined by homogeneous polynomials. Then, there is a finite semi-algebraic covering of S=∪ri=1Si, some formspi,j(x)∈R[x], some homogeneous matrix polynomials Ai,j(x)∈R[x]m×m such that,
∀x∈Si, A(x) =
ri
X
j=1
pi,j(x)ATi,j(x)Ai,j(x) with the condition thatpi,j(x)≥0 for allx∈Si.
Proof. We may perform the same proof as in Theorem 4.1.
As an example of a case whereA(x) is not definite, we consider a general matrix polynomial of degree at most 2 in a single (non homogeneous) variablex.
Example 5.5. Assume thatA(x) is psd in the neighbourhood of 0+.
We assume moreover that A(0) is not positive definite otherwise we are done by 5.4 or by a domination argument. Up to a base change, we may assume that A(0) =
1 0 0 0
. Then, let us write A(x) =
1 0 0 0
+x
a1 b1
b1 c1
+x2
a2 b2
b2 c2
=
1 +a1x+a2x2 b1x+b2x2 b1x+b2x2 c1x+c2x2
SinceA(x) is psd at 0+, we havec1≥0.
* Ifc1>0, then the domination is satisfied and we get the following certificate A(x) = (1−µx−αx2)
1 0 0 0
+x
a1+µ b1
b1 c1
+x2
a2+α b2
b2 c2+βc1
whereα, β, µ are positive real numbers chosen such that the constant ma- trices of the identity are psd. Note that (1−µx−αx2) is obviously positive on a neighbourhood of 0+.
* If c1 = 0, then A(x) is psd on 0+ ifc2−b21 ≥0. If c2−b21 >0, then we have the certificate
A(x) = (1−αǫx)
1 0 0 0
+ 1−ǫ b1x b1x (b1−ǫ1x)2
!
+x2 a2+αǫ b2
b2 c2−1−ǫb21
!
whereǫandαare positive real numbers such that the last constant matrix in the identify is positive definite.
* Now, consider the case whenc1= 0 andc2−b21= 0. By positivity ofA(x) we havea1c2−2b1b2≥0.
The case when b1 = 0 is trivial since we must have b2 = 0 and the certificate follows. Ifb16= 0, then we get the conditiona1b1−2b2≥0.
First assume thata1b1−2b2>0. We may write A(x) =
x
a1−2bb12 +x2
a2−
b2
b1
2 1 0 0 0
+
1 +bb2
1x2
(b1x) 1 + bb2
1x (b1x)
1 +bb2
1x
(b1x)2
* In the remaining case whenc1 = 0, c2−b21 = 0, b1 6= 0, a1b1−2b2 = 0, the positivity ofA(x) at 0+says thata2b21−b22≥0. The desired certificate follows from the identity given in the previous case.
This inspection of the most elementary situation leads to conjecture that any matrix polynomial in a single variable admits a local certificate of positivity.
We give a proof of this fact, although we do not give explicit formulas depending on the entries as in the previous worked example.
Theorem 5.6. LetM ∈R[x]n×n be a symmetric matrix polynomial whose entries are polynomials in a single variable x. Assume that M is psd on a neighbourhood of 0+.
Then,M admits a semi-certificate of positivity at0+.
Proof. If M(0) is invertible, then we are done by 4.1. Hence, from now on, we assume that det(M(0)) = 0.
Let us consider the Smith normal form ofM : M =EDF
where E and F are invertible in R[x]n×n andD = diag(d1, . . . , dn) is diagonal in R[x]n×n with diagonal entries (d1, . . . , dn). By changing M to (FT)−1M F−1, we may assume thatM =ED. Moreover, ifdi≡0 then we may argue by induction on the size ofM, hence we will assume in the following thatdi≥0 at 0+and vanishes only at 0.
The matrixM has the form
M =
d1e1,1 d2e1,2 . . . dne1,n
d2e1,2 d2e2,2 . . . dne2,n
... ...
dne1,n dne2,n . . . dnen,n
Hence the matrixE has the form
E=
e1,1 e1,2 . . . e1,n d2
d1e1,2 e2,2 . . . e2,n
... . .. . .. ...
... . .. . .. ...
dn
d1e1,n dn
d2e2,n . . . ddn
n−1e2,n en,n
First, let us introduce some notations.
Define by induction the following sequence of integers :
Letk0= 1 and setki to be the first integer j > ki−1 such that ddjj
−1(0) = 0. It yields an increasing sequence of integers :
1 =k0< k1< . . . < kr≤n=kr+1−1.
We divide the matrixM = (mk,l)1≤k,l≤n into block matrices : M = (Mi,j)0≤i,j≤r, where we setMi,j= (m(i,j)k,l )1≤k≤ki+1−ki,1≤l≤kj+1−kj withm(i,j)k,l =mk+ki−1,l+kj−1. Likewise, we divideE= (Ei,j)0≤i,j≤r into similar block matrices.
Let us define the block matrixMi = (Rk,l)0≤k,l≤r such thatRk,l=Mk,l when (1≤k≤i andl=i) or (1≤l≤i andk=i) andRk,l= 0 otherwise. Namely
Mi=
0 . . . 0 M1,i 0 . . . 0 ... ... ... ... ... 0 . . . 0 Mi−1,i ... ... Mi,1 . . . Mi,i−1 Mi,i ... ... 0 . . . 0 . . . 0
... ... ...
0 . . . 0 . . . 0
.
Let us write now
M =dk0M0+dk1M1+. . .+dkrMr.
Observe then by construction thatE(0) is block upper triangular. SinceE(0) is invertible we get thatEi,i(0) is invertible for eachi= 0, . . . , r.
Moreover, we have
(7) det(Mi,i) = det(Ei,i)×
ki+1−1
Y
j=ki
dj
By definition of theki’s, it implies that det(d1
kiMi,i) does not vanish at 0 . Hence we may derive a kind of domination condition for the matrixM at 0+. Namely, let us write
M = dk0M0−dk1×µ1×Idk0,k1−. . .−dkr×µr×Idk0,k1
+dk1(M1+µ1Idk0,k1)−dk2×µ2×Idk1,k2−. . .−dkr×µr×Idk1,k2
...
+dkr(Mr+µrIdk0,kr)
= dk0N0+dk1N1+. . .+dkrNr
where theµi’s are positive real numbers and Idp,q is the diagonal matrix inRn×n whose i-th diagonal entry is +1 ifp≤i ≤q−1 and 0 otherwise. Because of (7), we know that we can choose theµi’s such that the matrix polynomialsNi are psd on a neighbourhood of 0+.
Then, using the same argument as in Theorem 4.1, we complete the construction
of a semi-certificate.
The result is no more true with more than one variable : Proposition 5.7. Let
M(x, y) =
1 +x2−y2 −x
−x y2
.
Then, M(x, y)is psd for x2≥y2≥1 andx2 ≤y2≤1 althought it does not admit a certificate in a neighbourhood of (1+,1+).
Proof. First note that det(M(x, y)) = (y2−1)(x2−y2) to conclude to the positivity domain ofM(x, y).
For convenience, let us consider the following change of variables y = 1 +Y and x= Y +H. We also perform a base change to “simplify” the expression of M(0,0) =
1 −1
−1 1
. LetP =
1 −1
1 1
andQ=
1 1
−1 1
. Then, P×M(0,0)×Q=
4 0 0 0
and thus let us introduce the following matrix N(X, Y) = P×M(x, y)×Q
=
4 + 4H+ 4Y +H2+Y2+ 2HY 2H−2Y +H2−Y2+ 2HY 2H−2Y +H2−Y2+ 2HY H2+Y2+ 2HY
The matrixN(X, Y) is psd at an orthant-neighbourhood (0+,0+) with respect to the variables (H, Y). Let us assume that it has a semi-certificate of positivity.
Namely :
(8) N(X, Y) =X
i
Fi
ǫ2i ǫi ǫi 1
+X
j
Uj2 UjVj
UjVj Vj2
+G 1 0
0 0
where all Fi’s andGare psd on an orthant-neighbourhood (0+,0+). The degrees of the (2,2)-entries shows that eachfi has only monomials of degree 2 and eachVj has monomials of degree 1. Hence :
Fi=aiH2+biY2+ciHY Uj=γj+µjH+νjY Vj =αjH+βjY
G=a′H2+b′Y2+c′Y H+d′H+e′Y +f′ Identifying the (2,2)-entries, we get
(9)
X
i
ai+X
j
α2j = 1, X
i
bi+X
j
βj2= 1 X
i
ci+ 2X
j
αjβj= 2
Identifying the coefficients in H and Y of the (1,2)-entries and the constant coefficients of the (1,1)-entries yields :
(10)
X
j
αjγj = 2 X
j
βjγj=−2 X
j
γj2+f′= 4
Let us introduce ¯α = (αj), ¯β = (βj) and ¯γ = (γj), vectors of the standart euclidean space (whose dimension is equal to the number of indexesj).
After (10) and (9), we get
¯
γ2≤4, α¯2≤1 and α¯·γ¯= 2.
By the Cauchy- Schwartz case of equality, we get ¯γ2 = 4 and ¯α2 = 1. Thus, f′ = 0 andP
iai= 0 which means that for all i,ai = 0. Moreover, the vectors ¯α and ¯γ must be colinear and hence ¯α=12γ.¯
Likewiseβ =−12γandP
ici= 4. Moreover, for alliwe havebi= 0. And hence we may assume that there is only one indexiand we may write
(11) N(X, Y) = 4HY
ǫ2 ǫ ǫ 1
+X
j
Uj2 UjVj
UjVj Vj2
+G 1 0
0 0
. Identifying the (1,2)-entries, and setting ¯µ= (µj) and ¯ν = (νj) we have
(12)
1
2¯γ·µ¯= 1 1
2¯γ·ν¯= 1
¯ γ
2 ·(¯µ−¯ν) + 4ǫ= 2
We readily deduce thatǫ= 12. Identifying the (1,1)-entry, we have furthermore
(13)
2 ¯γ·µ¯+d′ = 4, 2 ¯γ·ν¯+e′= 4, 2 ¯µ·ν¯+c′+ 1 = 2,
¯
µ2+a′= 1,
¯
ν2+b′= 1,
By (12) and (13), we immediately haved′=e′= 0. Moreover, we get also
¯
γ·µ¯= 2, µ¯2≤1 and ¯γ2= 4.
By the Cauchy-Schwartz case of equality, we deduce thata′= 0 and ¯µ=12¯γ.
Likewiseb′ = 0 and ¯ν =12¯γ.
Then,G=c′HY where necessarily c′ ≥0. To conclude, it suffices to note that 2 ¯µ·ν¯+c′+ 1 = 2 gives 2 +c′+ 1 = 2, a contradiction.
Remember that any psd biquadratic form in dimension 2 is a sum of squares, and hence admits a semi-certificate. This example shows that thelocalcounterpart is no more true.
In general we may mention the following result. Although completely elementary, it looks very much like the ones we can find in [7] or [8] for hermitian squares or more classical sums of squares.
Proposition 5.8. Let A(x)∈R[x]m×m be a symmetric matrix polynomial. Then, A(x) is psd if and only if for all realǫ >0the matrix polynomial
Aǫ(x) =A(x) +ǫ Xn i=1
x2i
! Idm
admits a semi-certificate.
Of course the complexity of the semi-certificate may increaes asǫgoes to zero.
Remark 5.9. The complexity of a semi-certificate shall be measured by the size of the semi-algebraic partitions and also by the number of squares. Here the degrees of the polynomials appearing in a (homogeneous) semi-certificate are bounded by the degree of the given matrix polynomial.
Remark 5.10.By [6] with respect to the (compact) unit sphere{Pn
i=1x2i = 1}, then we obtain for M a Positivestellensatz which is no more homogeneous and where the degrees are no more bounded.
Remark 5.11. Consider a semi-certificate, and a compact semi-algebraic subsetSi
of the given partition. If we assume that the desciption ofSi is archimedean, then we may deduce a “true” algebraic certificate without denominators for the matrix polynomial onSi, using for instance Schm¨udgen or Putinar certificates.
6. Around the Choi counterexample
Let us consider the counterexample of a biqudratic psd non sum of squares given in [1] :
x2+ 2z2 −xy −xz
−xy y2+ 2x2 −yz
−xz −yz z2+ 2y2
Altough it is not a sum of squares, it admits a semi-certificate. Indeed, when
|x| ≥ |z|, it can be decomposed as
x2 −xy −xz
−xy y2 yz
−xz yz z2
+ 2
z2 0 0
0 x2 −yz 0 −yz y2
= (−x, y, z)T(−x, y, z) + 2(0,−z, y)T(0,−z, y)
+2z2
1 0 0 0 0 0 0 0 0
+ 2(x2−z2)
0 0 0 0 1 0 0 0 0
And by symmetry, we deduce an analogeous certificate when |z| ≥ |x|.
Let us consider a variation around this example ; in the following proposition, the Choi counterexample is justM1:
Proposition 6.1. Let
Mλ=
x2+ (λ+ 1)z2 −xy −xz
−xy y2+ (λ+ 1)x2 −yz
−xz −yz z2+ (λ+ 1)y2
whereλ∈R.
Then, Mλ is never a sum of squares and it is psd if and only if λ∈ [0,+∞[.
Moreover, det(M0) is a psd non sos polynomial andMλ admits a semi-certificate of positivity for any λ∈]0,+∞[.
Proof. The exact argument as in [1] works to show that any Mλ is not a sum of squares. We may reproduce it for convenience. We translate our ramewok into the language of biforms, setting
fMλ = (x2+(λ+1)z2)S2−2xyST−2xzSU+(y2+(λ+1)x2)T2−2yzT U+(z2+(λ+1)y2)U2 Assume thatfMλ =P
fi2 wherefi is a bilinear form. Since there is no monomials x2U2, y2S2, z2T2 in fMλ, there is no such monomials in each fi2, and hence no monomialsxU, yS norzT in each fi. We may write fi=gi+hi weregi depends only in the monomialsxS,yT,zU andhi depends only in the monomialsxT,yU, zS.
Then,fMλ =P
(gi+hi)2 shows that X
i
g2i =x2S2+y2T2+z2U2−2(xyST+yzT U+xzSU),
a contradiction since the right hand side of the equality takes negative values for x=y=z=S=T =U = 1.
The fact thatM0is psd and not a sum of squares may also be shown directly by using the following consequence of the Cauchy-Binet formula.
Lemma 6.2. Let M(x) be a symmetric matrix polynomial. If M is a sum of squares, then its determinant is also a sum of squares. 1.
Proof. We first state the Cauchy-Binet formula.
Given matricesA∈Rm×sandB∈Rs×m, the Cauchy-Binet formula states that det(AB) =X
S
det(AS) det(BS)
where S ranges over all the subsets of {1, . . . s} with m elements, and AS (re- spectivelyBS) denotes the matrix inRm×m whose columns are the columns ofA (respectively whose rows are the rows ofB) with index fromS.
IfM(x) is a sum of squares, then it can be writtenM(x) =AT(x)A(x) for some matrix polynomialA(x)∈R[x]s×m. Then,
det(M(x)) = P
S det(A(x)T)S det(A(x)S)
= P
S det(MS(x)T) det(MS)
= P
S( det(MS))2.
1The Cauchy-Binet formula shows even more : ifM is a sum of squares, thenallits principal minors are sums of squares.
Note thatM0 has as a determinant :
det(M0) =x4y2+y4z2+z4x2−3x2y2z2
a variation of the celebrated Motzkin polynomial which is psd and not a sum of squares !
Thus, the matrix polynomial M0, which is clearly psd since all its principal minors are psd, cannot be a sum of squares since its determinant is not !
Thus,M0(and thus allMλ forλ≥0) are psd matrix polynomial. Moreover, det(Mλ(1,1,1)) =λ(λ+ 3)2
which shows thatMλis not psd for smallλ <0 and hence for allλ <0.
Let us study now the existence of semi-certificate of positivity.
Forλ >0, the set of real singular points of det(Mλ) have projective coordinates is [1 : 0 : 0], [0 : 1 : 0] and [0 : 0 : 1] (they correspond to the points where the non negative polynomial det(Mλ) vanishes). The identity
Mλ=
x2 −xy −xz
−xy y2 yz
−xz yz z2
+
λz2 0 0
0 4zλ2 −2yz 0 −2yz λy2
+1
λ(λx2−4z2)
0 0 0 0 1 0 0 0 0
gives a semi-certificate at the neighbourhood of [1 : 0 : 0]. We may also proceed likewise at the neighbourhood of [0 : 1 : 0] and [0 : 0 : 1].
Elsewhere the matrix Mλ is positive definite so that we can argue as in the proof of Theorem 4.1 and obtain a finite semi-algebraic covering, and hence a semi- certificate.
Whereas, the matrice M0 has a lot more singular points, namely : [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1], [1 : 1 : 1], [1 : 1 :−1], [1 :−1 : 1], [1 :−1 : 1].
So, the problem which remains is to find a certificate of positivity forM0at the points [1 : 1 : 1], [1 : 1 :−1], [1 :−1 : 1], [1 :−1 : 1]. It does not appear trivial.
For instance, the domination condition is not satisfied here. Another natural idea would be to consider :
M0 =
x2+z2 −xy −xz
−xy y2+x2 −yz
−xz −yz z2+y2
=
0 0 0
0 y2 −yz 0 −yz y2
+
y2 −xy 0
−xy x2 0
0 0 0
+
x2+z2−y2 0 −xz
0 0 0
−xz 0 y2
But, setting z= 1 in the last matrix, we get a matrix polynomial which is psd at an orthant-neighbourhood of (1+,1+) with respect to the variables (x, y), but which does not admit any semi-certificate as shown by Proposition 5.7.
Nevertheless, let us see in the following how it is possible to get a semi-certificate in some orthant-neighbourhood of [1 : 1 : 1], and hence a partial result.
